A particle is projected with an angle of projection $\theta$ to the horizontal line passing through the points $( P , Q )$ and $( Q , P )$ referred to horizontal and vertical axes (can be treated as $x$-axis and $y$-axis respectively).
The angle of projection can be given by
$\tan ^{-1}\left[\frac{ P ^2+ PQ + Q ^2}{ PQ }\right]$
$\tan ^{-1}\left[\frac{ P ^2+ Q ^2- PQ }{ PQ }\right]$
$\tan ^{-1}\left[\frac{ P ^2+ Q ^2}{2 PQ }\right]$
$\sin ^{-1}\left[\frac{ P ^2+ Q ^2+ PQ }{2 PQ }\right]$
The angle of projection at which the horizontal range and maximum height of projectile are equal is
If T is the total time of flight, $h$ is the maximum height $ \& R$ is the range for horizontal motion, the $x$ and $y$ co-ordinates of projectile motion and time $t$ are related as
Column $-I$ Angle of projection |
Column $-II$ |
$A.$ $\theta \, = \,{45^o}$ | $1.$ $\frac{{{K_h}}}{{{K_i}}} = \frac{1}{4}$ |
$B.$ $\theta \, = \,{60^o}$ | $2.$ $\frac{{g{T^2}}}{R} = 8$ |
$C.$ $\theta \, = \,{30^o}$ | $3.$ $\frac{R}{H} = 4\sqrt 3 $ |
$D.$ $\theta \, = \,{\tan ^{ - 1}}\,4$ | $4.$ $\frac{R}{H} = 4$ |
$K_i :$ initial kinetic energy
$K_h :$ kinetic energy at the highest point
A body is projected at such an angle that the horizontal range is three times the greatest height. The angle of projection is
Aball is thrown from a point on ground at some angle of projection. At the same time a bird starts from a point directly above this point of projection at a height $h$ horizontally with speed $u$. Given that in its flight ball just touches the bird at one point. Find the distance on ground where ball strikes